Question 6.6.3: The trapezoidal profile requirements for a specific applicatio......
The trapezoidal profile requirements for a specific application are given in the following table, along with the load and motor data. Determine the motor and amplifier requirements.
|
t_{1}=0.2\;{\mathrm{s}},t_{2}=0.4\;{\mathrm{s}},t_{f}=0.6\;{\mathrm{s}} |
Cycle times |
Profile data
|
|
Displacement \theta_{L f}=10\pi{\mathrm{~rad}} Reduction ratio N =2 |
Inertia I_{L}=4\times10^{-3}\,\mathrm{kg}\cdot\mathrm{m}^{2} Torque {{T}}_{L}=0.1\,\mathrm{N}\cdot\mathrm{m} |
Load data |
|
Torque constant K_{T}=0.3\;\mathrm{N}\cdot\mathrm{m/A} Damping c = 0
1.56\times10^{-3} s and 0.043 s |
Resistance R = 2 Ω Inductance L=3\times10^{-3}\,\mathrm{H} Inertia I_{m}=10^{-3}\,\mathrm{kg}\cdot\mathrm{m}^{2} Time constants |
Motor data |
The total inertia I is the sum of the motor inertia and the reflected load inertia. Thus,
I=I_{m}+{\frac{I_{L}}{N^{2}}}=10^{-3}+{\frac{4\times10^{-3}}{2^{2}}}=2\times10^{-3}~{\mathrm{kg}}\cdot{\mathrm{m}}^{2}Because the reduction ratio is N = 2, the required motor displacement is N\theta_{L f} = 2(10π) = 20π rad, and the load torque as felt at the motor shaft is T_{d}= T_{L}/N = 0.1/2 = 0.05 N · m.
The motor’s energy consumption per cycle is found from (6.6.13).
E=\frac{R}{K_{T}^{2}}\left(\frac{2I^{2}\theta_{f}^{2}}{t_{1}t_{2}^{2}}+T_{d}^{2}t_{f}\right) (6.6.13)
E={\frac{2}{(0.3)^{2}}}\left[{\frac{2(4\times10^{-6})(20\pi)^{2}}{0.2(0.4)^{2}}}+(0.05)^{2}0.6\right]=22\,{\mathrm{J/cycle}}The power consumption is 22/t_{f} = 37 J/s, or 37 W.
Equation (6.6.12) shows that the maximum speed for the trapezoidal profile is
\omega_{\mathrm{max}}={\frac{\theta_{f}}{t_{2}}} = 50π rad/s
which is 50π (60)/(2π) = 1500 rpm. So the motor’s maximum permissible speed must be greater than 1500 rpm.
The rms torque is found from (6.6.17) to be
T_{\mathrm{rms}}=\sqrt{\frac{2I^{2}\theta_{f}^{2}}{t_{f}t_{1}t_{2}^{2}}+T_{d}^{2}} (6.6.17)
T_{\mathrm{rms}}={\sqrt{\frac{2(4\times10^{-6})(20\pi)^{2}}{0.6(0.2)(0.4)^{2}}+(0.05)^{2}}}=1.28~\mathrm{N}\cdot\mathrm{m}Use (6.6.16) to compute the maximum required torque.
T_{\mathrm{max}}=I\alpha_{\mathrm{max}}+T_{d}=I\frac{\omega_{\mathrm{max}}}{t_{1}}+T_{d}=I\frac{\theta_{f}}{t_{1}t_{2}}+T_{d} (6.6.16)
T_{\mathrm{max}}={\frac{2\times10^{-3}(20\pi)}{0.2(0.4)}}+0.05=1.57+0.05=1.62\,\mathrm{N}\cdot{\mathrm{m}}Note that the load torque contributes little to T_{\mathrm{max}}. Most of the required torque is needed to accelerate the inertia.
The system time constants are obtained from the roots of (6.5.13), which are s = −643 and s = −23.3. The system must be fast enough to respond to the profile command. Its largest time constant, 1/23.3 = 0.043 s, is less than one-fourth of the ramp time t_{\mathrm{1}} = 0.2, so the system is fast enough.
The amplifier requirements are calculated as follows. Note that because the motor data is given in SI units, K_{b}{=}\ K_{T} = 0.3. From (6.6.18), (6.6.19), and (6.6.20),
i_{\mathrm{max}}={\frac{T_{\mathrm{max}}}{K_{T}}} (6.6.18)
i_{\mathrm{rms}}={\frac{T_{\mathrm{rms}}}{K_{T}}} (6.6.19)
v_{\mathrm{max}}=R i_{\mathrm{max}}+K_{b}\omega_{\mathrm{max}} (6.6.20)
i_{\mathrm{max}}={\frac{1.62}{0.3}}=5.4\,\mathrm{A}i_{\mathrm{rms}}={\frac{1.28}{0.3}}=4.27\,\mathrm{A}
v_{\mathrm{max}}=2(5.4)+0.3(50\pi)=10.8+47.1=57.9\,\mathrm{V}
Note that most of the required voltage is needed to oppose the back emf.